Download 6. Magnetic Cooling - Particle Physics

Statistical and Low Temperature Physics (PHYS393)
6. Magnetic Cooling
Kai Hock
2013 - 2014
University of Liverpool
Learning Aims: You will learn to
State the formula for energy levels in a spin 1/2 paramagnetic
salt.
Sketch and explain the temperature graphs for energy, heat
capacity and entropy of the spin 1/2 salt.
State the distribution of spin 1/2 ions among these levels.
Derive the formula for total energy.
Use the entropy graph to explain magnetic cooling.
Explain how to find the heat of magnetisation, the cooling
power, and the final cooling temperature.
Explain what limits the lowest temperature that can be
achieved. Explain how nuclear cooling overcomes this problem.
1
Paramagnetic salts
2
Paramagnetic salts.
Heat capacities tend to increase with temperature, as we have
seen for electrons in metals and phonons in solids. A
paramagnetic salt, however, has a heat capacity that goes up
to a peak, comes down, then goes up again.
J.W. Stout, W.B. Hadley, J. Chem. Phys. 40, 55 (1964)
This unusual peak is called the Schottky anomaly.
3
Schottky anomaly
The Schottky anomaly happens for paramagnetic materials in a
magnetic field. At very low temperatures, there is little
contribution from electrons and phonons (lattice vibration).
Consider a material with ions that have spin 1/2. This is called
a spin 1/2 paragmagnet. When a magnetic field B is applied,
the energy level of each ion split into two.
The energy levels are
ε = −µ.B = −µB B or + µB B
4
Lets see if we can “deduce” the Schottky anomally. At 0 K, all
electrons are at the ground state, so the total energy U is
−N µB B, where N is the number of the ions.
At high temperatures, the thermal energy the electrons is much
larger than the difference between the two magnetic energy
levels. Then the electrons are equally likely to be in −µB B or
+µB B, so U is zero
A sketch of U against T would look like this.
5
Heat capacity
In principle, we can now find the heat capacity C using
dU
C=
.
dT
However, the gradient of U neat 0 K is not known.
According to one form of the Third Law of Thermodynamics:
”The entropy of a system approaches a constant value as the
temperature approaches zero.”
This means that change in entropy dS approaches zero. Since
dQ
dS =
,
T
so
dU
dQ
dS
C=
=
=T
→ 0.
dT
dT
dT
6
Now that we know the gradient is zero at both 0 K and high T
we can sketch the heat capacity, which is the gradient of U .
We have just deduced the Schottky anomaly.
7
The energy can be obtained quantitatively using the Boltzmann
distribution of the number of spin 1/2 ions in the energy levels
1 and 2:
µB B
n1 = A exp +
kB T
!
!
µB B
and n2 = A exp −
.
kB T
Using the total number
N = n1 + n2,
we can solve for A and find the total energy
U = n1(−µB B) + n2(µB B).
The answer is
µB B
U = −N µB B tanh +
kB T
8
!
The entropy.
Next, we shall deduce qualitatively the graph for the entropy. It
looks like this:
We need to show that at low T, it tends to zero. And at high
T, it tends to N kB ln 2.
Recall the formula for entropy: S = kB ln Ω.
9
Behaviour of the entropy
At low temperature, most atoms would be in the ground state.
There is only 1 way to arrange this. So the entropy would tend
to S = kB ln 1 = 0.
At high temperature, the difference between the energy levels
would be small compared with the kB T in exp(−/kB T ).
The atom is equally likely to be in either level. There are 2
possible arrangements for each of the N atoms - 2N in total.
So the entropy would tend to S = kB ln 2N = N kB ln 2.
10
The Principle of Magnetic Cooling
11
The Principle of Magnetic Cooling
Magnetic cooling can ultimately reach a temperature of
microKelvins.
To understand the principles, we start with the paramagnetic
salt, which can go down to milliKelvins. This salt contains ions
with magnetic moments coming from their electrons.
At milliKelvin temperatures, the magnetic disorder entropy
(about 1 J/mol) is large compared to all other entropies, such
as lattice and conduction electron entropies, which may be
neglected.
We have seen the properties of a paramagnetic salt. The
entropy is a function of the magnetic field applied. We shall
look at this and see how it can help us understand magnetic
cooling.
12
The Principle of Magnetic Cooling
This graph show the entropy against temperature for a
commonly used paramagnetic salt.
D.S. Betts: An Introduction to Millikelvin Technology (Cambridge University Press,
Cambridge 1989)
We start by familiarising ourselves with the various features of
the graph.
13
The Principle of Magnetic Cooling
The symbol B is the magnetic field. We start with the graph
for zero or low magnetic field.
14
The Principle of Magnetic Cooling
At high temperature, the entropy approaches a constant
because it becomes equally likely to be at any of the magnetic
energy levels.
15
The Principle of Magnetic Cooling
At low temperature, the entropy goes to zero because all
particles fall to the lowest level.
16
The Principle of Magnetic Cooling
Next, suppose the magnetic field is increased, say from 0 T to
0.1 T. The spacing between energy level would increase.
Then it becomes more likely for a particle to be at the lower
level. So the entropy would fall.
17
The Principle of Magnetic Cooling
Point A: To understand how to use the magnetic property for
cooling, suppose we start with a temperature at point A, and
with a low magnetic field..
The salt is placed in contact with a precooling bath. This can
either be a helium bath, or a dilution refrigerator.
18
The Principle of Magnetic Cooling
Point B: A magnetic field is then applied. This is done
isothermally. The entropy falls to point B at constant
temperature.
This process performs magnetic ”work” on the salt, which is
converted to heat (like compressing a gas). This heat would be
absorbed by the precooling bath.
19
The Principle of Magnetic Cooling
The salt is then thermally isolated from the precooling bath
(e.g. by using a heat switch).
Point C: Demagnetisation now takes place adiabatically (so
entropy is constant). The magnetic field is reduced to a very
small value. The temperature falls to C.
20
How does it work?
We have seen how the cooling takes place using
thermodynamics. Let us now see how this takes place
physically.
The ions in the salt have magnetic dipole moments. Normally,
half of the dipoles are spin up, and the other half are spin down.
If a strong magnetic field is applied, the energy levels will split
into two.
Dipoles in the direction of the field will have lower energy, and
dipoles in the opposite direction higher energy.
21
Precooling
Remember that a helium bath or a dilution refrigerator is
cooling the salt at the same time.
This would remove energy from the higher energy atoms, so
that they fall into the lower energy state.
This is the ”precooling.”
22
An adiabatic, constant entropy change.
Then, using a heat switch, the salt can be thermally
disconnected from the precooling bath.
The magnetic field is now slowly reduced.
The lower energy level, which contains most of the atoms, is
then forced to increase in energy.
This energy has to come from the surrounding. So the salt
cools down.
23
The Principle of Magnetic Cooling
Point C: After some time, the salt would warm up because of
heat leak from the surroundings, since the insulation is not
perfect. The temperture returns to point A along the curve
from C.
The temperature cannot be maintained, unlike the dilution
refrigerator. This is called a ”one-shot” technique.
24
Thermodynamics of Magnetic Cooling
25
Thermodynamics of Magnetic Cooling
With the help of the entropy graph, we can calculate the heat
Q and temperature T in magnetic cooling.
We can divide the cycle into the following stages:
1. Isothermal magnetisation: We shall find the heat given out
by the salt.
2. Adiabatic demagnetisation: We shall find the lowest
temperature reached.
3. Warming up: We shall determine the cooling power.
26
Thermodynamics of Magnetic Cooling
Isothermal magnetisation takes place from A to B on the graph
at the start. Since dQ = T dS, the heat given out is
Q=
Z A
B
T dS.
This is just the area of the rectangle ABDE.
The heat released is usually a few J/mol of the refrigerant (the
salt), so it can easily be absorbed by an evaporating helium
bath or a dilution refrigerator.
27
Thermodynamics of Magnetic Cooling
Adiabatic demagnetisation takes place from B to C on the
graph.
Later on, we shall derive the formula for magnetic entropy. In
that formula, we shall see that the entropy is a function of B/T .
For now, a quick way to understand this is to look at the
Boltzmann distribution, exp(−µB B/kB T ). This is indeed a
function of B/T .
28
Thermodynamics of Magnetic Cooling
Since the entropy should depend on exp(−µB B/kB T ), we would
expect the entropy to be a function of B/T as well.
This gives us a quick way to find the coldest temperature
reached in this demagnetisation step.
From the graph, we see that the entropy is a simple function of
T and B. If T increases, entropy increases. If B increases,
entropy decreases.
In the adiabatic process, the entropy is a constant. Since it is a
function of B/T , then B/T must also be constant. This is just
the equation we need:
B
= constant
T
29
Thermodynamics of Magnetic Cooling
Point B: Suppose we start at temperature Ti and field Bi. If
this is point B on the graph, then Bi = 0.1T and Ti = 100mK
Point F: We then reduce the field to a smaller value Bf . Let Tf
be the new temperature.
The equation B/T = constant means that
Bf
Tf
=
30
Bi
Ti
Thermodynamics of Magnetic Cooling
The new temperature is
Ti
Tf =
Bf
Bi
Clearly, we can make Tf very small by reducing the magnetic
field Bf to a very small value.
But what if we reduce the magnetic field Bf to zero? Surely
the temperature Tf would not go to zero. Something must
happen to limit the lowest temperature that we can reach.
Indeed, this is the case. When the temperature is sufficiently
low, the effects of the magnetic fields from neighbouring ions
of the salt become important.
31
Interacting Magnetic Dipoles
32
Thermodynamics of Magnetic Cooling
When the temperature is sufficiently low, the mutual
interaction would tend to align all magnetic dipoles in the same
direction. When this happens, the entropy falls to zero, and the
magnetic cooling would stop.
So the mutual interaction limits the lowest temperature that
can be achieved using this method. At very low temperatures,
the equation would have to be modified to take into account
this mutual interaction. We shall come back to this.
33
Thermodynamics of Magnetic Cooling
Point C: We can now see that the graph is misleading. We do
not actually demagnetise to a field at point C, which is zero.
Rather, we would demagnetise to a very small field, close to C.
To make things simple, we shall still refer to C for the end
point of the demagnetisation. Then warming up starts.
34
Thermodynamics of Magnetic Cooling
Remember that the salt is thermally isolated during the
demagnetisation. After reaching the lowest temperature at C,
the salt remains isolated. We hope that it would stay cold for
as long as possible.
But because insulation is not perfect, the salt starts warming
up slowly. Since the magnetic field B is fixed, the temperature
and entropy would follow the curve and eventually reach the
starting temperature at A.
35
Thermodynamics of Magnetic Cooling
As it warms up, the heat absorbed by the salt can be obtained
from the entropy using the same formula as before
Q=
Z A
C
T dS
We need to integrate along the curve from C to A. So the heat
absorbed is given by the shaded region on the graph.
36
Thermodynamics of Magnetic Cooling
The heat absorbed in warming up also gives the cooling power.
If the salt can absorb more of the heat that leaks in through
the insulation, then it would be able to remain cold for a longer
period of time.
Note that this is a different definition from before. For the
dilution refrigerator, the cooling power Q̇ is the rate at which
heat is absorbed.
The cooling power Q for the magnetic refrigerator is the
total heat absorbed.
37
Interaction between magnetic dipoles
We have seen that the interaction between magnetic dipoles of
the ions in the salt sets a lower limit to the temperature that
can be reached by demagnetisation.
This also means that the formula for the lowest temperature
Ti
Bf
Bi
would not be accurate at very low temperatures. It can be
modified to the following form:
Tf =
Ti
Bf2 + b2
Bi
r
Tf =
38
Interaction between magnetic dipoles
Lets try and understand this formula physically.
Ti
Bf2 + b2
Bi
r
Tf =
We see that when Bf is reduced to zero,
Tf =
Ti
b.
Bi
If we compare this with the original form of
Tf =
Ti
Bf
Bi
we see that b corresponds to Bf . This makes sense if we think
of b as the field that remains after the applied magnetic field
has been reduced to zero.
39
Interaction between magnetic dipoles
When the applied magnetic field is reduced to zero, there is
indeed a remaining field. That would be the resultant field
from the neighbouring magnetic dipoles.
We are looking at a temperature Tc at which the effect of this
interaction becomes important. This means that kB Tc is
comparable to the interaction energy
εd = µb.
Tc is called the ordering temperature. It is the temperature
below which the neighbouring fields become strong enough to
align the dipoles. We may define
kB Tc = µb.
40
Magnetic Refrigerators
41
Magnetic Refrigerators
The performance of a magnetic refrigerator is mainly
determined by:
- the starting magnetic field and temperature,
- the heat leaks, and
- the paramagnetic salt that is used.
Typical starting conditions are 0.1 to 1 T and 0.1 to 1 K.
These are fairly easy to achieve nowadays.
There a few properties of a paramagnetic salt that are
desirable:
- low ordering temperature to reach low temperatures,
- large specific heat to absorb more heat before warming up
42
Magnetic Refrigerators
The following are paramagnetic salts that have been used:
The last one, CMN, has the lowest ordering temperature. This
means it can potentially reach the lowest temperature before
the interaction between magnetic dipoles become important.
CMN has been used extensively and could reach 2 mK.
43
Magnetic Refrigerators
The follow graphs show the entropies of the actual salts.
Pobell, Matter and methods at low temperatures (2007)
44
Magnetic Refrigerators
For example, look at the 2 graphs for the salt CMN:
The one to the left is for zero magnetic field.
The one to the right is for 2 T.
They look very similar to the entropy graphs that are shown
earlier.
45
Magnetic Refrigerators
These are examples of actual magnetic refrigerators that have
been built.
Pobell, Matter and methods at low temperatures (2007)
Notice the heat switch near the top. They are usually
connected to a dilution refrigerator. The paramagnetic salt
refrigerant is in the middle.
46
Magnetic Refrigerators
Magnetic refrigerators using paramagnetic salts are now largely
replaced by the dilution refrigerator, which can reach the same
temperatures.
However, they are still useful for small experiments and
satellites, where compact refrigerators are required. Examples
are in detectors for millimetre wave, X rays and dark matter.
47
Nuclear Refrigeration
48
Nuclear Refrigeration
We have so far looked at the use of the electronic magnetic
dipoles for cooling. This is limited to milliKelvin temperatures
by the interaction between the electronic dipoles.
It is possible to reach much lower temperatures if we use the
nuclear magnetic dipoles. The magnetic dipole moment of the
nucleus is much smaller than that of the electron. As a result,
the interaction between nuclear dipoles is much weaker.
49
Nuclear Refrigeration
To get some idea of the relative magnitudes, we look at the
unit for electronic dipole moment (Bohr magneton) and the
unit for nuclear dipole moment (nuclear magneton):
Bohr magneton, µB = 9.27 × 10−24 J/T
Nuclear magneton, µn = 5.05 × 10−27 J/T
The nuclear magneton is nearly 2000 times smaller. This gives
us an idea of how much smaller the nuclear magnetic moment
is.
If we use the nuclear magnetic dipole for cooling, we can reach
microKelvin temperatures because of the much smaller
interaction field. The ordering temperature for the nuclear
dipole can be as small as 0.1 µK.
For nuclear cooling, we can use metal as the refrigerant instead
of paramagnetic salts. Metal has the advantage of high thermal
conductivity.
50
Nuclear Refrigeration
Although the small nuclear moment offers the potential of
reaching much lower temperatures, it also requires much more
demanding conditions. The very small moment means that we
need very high starting magnetic fields, and very low starting
temperatures.
As an example, we look at copper. Copper is a ”work horse” of
nuclear refrigeration. For copper, we would typically need a
starting field of Bi = 8 T, and a starting temperature of
Ti = 10 mK. This is just to reduce the entropy by 9%.
From the earlier explanation on the principle of magnetic
cooling, we know that an lower entropy means that a lower
temperature can be reached during demagnetisation. It also
means a higher cooling power, since more heat can be
absorbed during warming up.
51
Nuclear Refrigeration
From the entropy graph, we can see that to reduce the entropy
further, even higher fields and lower temperatures would be
required.
The 8 T field already requires superconducting magnets, and
the 10 mK temperature would require a dilution refrigerator.
52
Nuclear Refrigeration
Why would a small magnetic moment require higher starting
field and lower starting temperature? Lets try and understand
this physically.
Consider the magnetic energy levels of copper. Applying a
magnetic field increases the spacing between levels. Because of
the small nuclear moment, the spacing would be small even for
a high starting field.
Pobell, Matter and methods at low temperatures (2007)
53
Nuclear Refrigeration
Because the spacing is small, we would get more particles at
the higher energy level according to the Boltzmann
distribution. This means higher entropy. In order to reduce
this, we need a lower starting temperature.
Pobell, Matter and methods at low temperatures (2007)
This is difficult. For copper, even at a starting temperature of
10 mK, there is still a substantial fraction of the nuclei at the
higher levels. The levels are just too close together because of
the small nuclear moment.
54
Nuclear Refrigeration
Fortunately, because the nuclear moment is small, the
interaction field is also small. This means that in the
demagnetisation step, it is possible to reduce the temperature
to a very small value.
In the case of the copper example, if we reduce the field by
1000 times to 8 mT, the temperature also falls by 1000 times
to 6 µK.
55
Nuclear Refrigeration
One disadvantage of very low temperatures in magnetic cooling
is that the cooling power becomes very small. The cooling
power is given by the shaded region in the graph.
The horizontal axis is temperature. So for low temperatures,
the horizontal size of the shaded area would also be small.
Since nuclear cooling is 1000 times colder than electronic
magnetic cooling, the cooling power is also 1000 times smaller.
56
Nuclear Refrigeration
This is a schematic diagram of a nuclear refrigerator. Notice
how the main components are connected together.
The refrigerant is a copper block at the centre.
Pobell, Matter and methods at low temperatures (2007)
57
Learning Aims: You should be able to
State the formula for energy levels in a spin 1/2 paramagnetic
salt.
Sketch and explain the temperature graphs for energy, heat
capacity and entropy of the spin 1/2 salt.
State the distribution of spin 1/2 ions among these levels.
Derive the formula for total energy.
Use the entropy graph to explain magnetic cooling.
Explain how to find the heat of magnetisation, the cooling
power, and the final cooling temperature.
Explain what limits the lowest temperature that can be
achieved. Explain how nuclear cooling overcomes this problem.
58
Worked Examples
59
Example 1
There is one mole of a spin 1/2 salt in a magnetic field of 2 T.
(i) Write down the formulae for the magnetic energy levels.
Calculate them.
(ii) Write down the formula for the distribution of number of
spin 1/2 ions in each level. Find the high temperature limit of
each formula.
(iii) Hence find the population in each energy level in the high
temperature limit.
60
Solution
(i) Formulae for the magnetic energy levels are
ε1 = −µB B, ε2 = µB B.
Substituting B = 2 T, we get
ε1 = −1.854 × 10−23 J, ε2 = 1.854 × 10−23 J.
(ii) The particles follow Boltzmann distribution:
!
!
µB B
µB B
n1 = A exp
, n2 = A exp −
.
kB T
kB T
At high temperatures, T is large and the arguments ε1/kB T and
ε2/kB T both tend to zero.
So the high temperature limits are:
n1 = A, n2 = A.
61
(iii) The total population of spin 1/2 ions is 1 mole.
Since the populations are equal in both levels, the population
of spin 1/2 ions in each level is 0.5 mole.
62
Example 2
There is one mole of a spin 1/2 salt in a magnetic field of 2 T.
(i) Write down the formula for the Boltzmann factor. Find the
Boltzmann factor at each energy level when temperature is 0.5
K.
(ii) Find the ratio of the population at the higher energy level
to the population at the lower energy level.
(iii) Find the population at each energy level in moles.
63
Solution
(i) The Boltzmann factors at levels 1 and 2 are :
µB B
exp
kB T
!
µB B
and exp −
kB T
!
respectively.
Substituting B = 2 T and T = 0.5 K, we find
µB B
kB T
exp
!
µ B
= 14.69 and exp − B
kB T
!
= 0.06809.
(ii) The populations are
!
n1 = A exp
!
µ B
µB B
, n2 = A exp − B
.
kB T
kB T
The ratio is
µB B
n1 : n2 = exp
kB T
64
!
µB B
: exp −
kB T
!
is the ratio of the Boltzmann factors.
So the answer is n1 : n2 = 14.69 : 0.06809.
(iii) The total population of spin 1/2 ions is 1 mole.
This is distributed between the two energy levels in the ratio
14.69 : 0.06809.
So the population in level 1 is
14.69
× 1 = 0.9954
14.69 + 0.06809
and the population in level 2 is
n1 =
0.06809
n2 =
× 1 = 0.004615
14.69 + 0.06809
65
Example 3
There is one mole of a spin 1/2 salt in a magnetic field of 2 T.
At first, the salt is at a high temperature where the populations
of spin 1/2 ions at each energy level are approximately equal.
(i) When temperature is lowered to 0.5 K, how much of the
population falls from higher to lower energy level? (Use results
from Example 2.)
(ii) When this happens, how much heat is given out? (Use
results from Example 1.)
66
Solution
(i) The total population is 1 mole. So at the high temperature,
there is 0.5 mole at each level:
n1 = 0.5, n2 = 0.5.
From Example 2, there are at 0.5 K the following:
n1 = 0.9954, n2 = 0.004615.
So the amount that falls from higher to lower level is:
(n2 at high temperature) - (n2 at 0.5 K) = 0.5 - 0.004615 =
0.4954 mole
(ii) From Example 1, the energies of each level are
ε1 = −1.854 × 10−23 J, ε2 = 1.854 × 10−23 J.
67
When a particle falls from higher to lower level, the heat it
gives out is ε2 − ε1.
0.4954 mole of spin 1/2 ions fall from higher to lower level.
This number of spin 1/2 ions is 0.4954 NA.
So the total heat given out is:
0.4954NA × (ε2 − ε1) = 11.06 J.
68
Example 4
This question follows from Example 3. The salt is now isolated
thermally, so that no heat can go out or come in.
(i) Write down the formula for the final temperature in
adiabatic demagnetisation, explaining all symbols used. The
magnetic field is then lowered from 2 T to 0.2 T. Find the new
temperature of the salt.
After some time, the temperature slowly increases because
heat leaks in.
(ii) The spin 1/2 ions move from lower to higher level until
populations are approximately equal again. Find the heat
absorbed.
69
Example 4
(i) Formula for the final temperature in adiabatic
demagnetisation is
Bf
Tf =
Ti,
Bi
where Ti is initial temperature, Tf is final temperature, Bi is
initial magnetic field and Bf is final magnetic field.
From Example 3, Ti = 0.5 K, Bi = 2 T.
Given Bf = 0.2 T. Substituting, we find
Tf =
0.2
× 0.5 = 0.05 K.
2
(ii) We know from Example 3 that 0.4954 mole of particles has
fallen from higher to lower energy.
70
When the salt warms up, the same amount would move back
to higher level. However, the energy spacing between levels is
now smaller, because the field has decreased.
The energy spacing is
ε2 − ε1 = µB B − (−µB B) = 2µB B.
This spacing is proportional to B.
The field has decreased from 2 T to 0.2 T. This is 10 times
smaller.
This means that the energy spacing is also 10 times smaller.
Therefore, when the particles move back to higher level, the
energy gain would also be 10 times smaller.
From Example 3, the heat given out was 11.06 J. The energy
gain now would be 10 times smaller than this.
So the answer is 11.06/10 = 1.106 J.
71